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Q.

2 mole of an ideal gas is expanded adiabatically and reversibly from 10 L to 80 L. The initial temperature of gas was 47°C. U for this process is ____ cal (Cv, of gas =  3 R/2).

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Detailed Solution

For adiabatic process, q = 0

 For gas, CV=32R,CP=CV+R=32R+R=52R and γ=CP/CV=53

 Now for adiabatic reversible process, TVγ-1 or, T2=T1V1V2γ1T2=T1V2V1γ-1=320108053-1 = 80 K Now, ΔU=W=nCVT2T1=2×32R×(80320)=1440 cal

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